Thread: Passing 2-D array to a Function

You really need to indent your code properly, e.g.,

Code:

#include <stdio.h>

int first(int(*q), int, int);
int second(int(*q)[4], int, int);

int main()
{
    int s[3][4] = {1, 2, 3, 4,
                   5, 6, 7, 8,
                   9, 10, 11, 12};
    first(s, 3, 4);
    second(s, 3, 4);
    return 0;
}

int first(int *q, int x, int y)
{
    int i, j;
    for (i = 0; i < x; i++)
    {
        for (j = 0; j < y; j++)
            printf("%d ", *(q+i*y+j));
        printf("\n");
    }
    printf("\n");
    return(*(q+i*y+j));
}

int second(int (*q)[4], int x, int y)
{
    int i, j, *p;
    for (i = 0; i < x; i++)
    {
        p = q + i;
        for (j = 0; j < y; j++)
            printf("%d ", *(p + j));
        printf("\n");
    }
    return(*(p+j));
}

Anyway, the warnings that you are getting likely have to do with the fact that s is an array of 3 arrays of 4 ints, so it is converted to a pointer to an array of 4 ints. However, in main, you pass it as an argument to a function that expects a pointer to an int:

Code:

first(s, 3, 4);

Then in second, you do pointer arithmetic with the pointer parameter that is a pointer to an array of 4 ints, but then assign the result to a pointer to an int instead:

Code:

p = q + i;

Originally Posted by Shankar k

would you please explain about the printf ("%d ", *(q+i*y+j)); statement.

The author decided to reinterpret the array of arrays as a single contiguous array of elements. The idea is basically to compute the index such that you get the equivalent of:

Code:

printf("%d ", s[i][j]);

where s is the array from main. Frankly, this code has overuse of pointer notation where array notation is appropriate.

warnings are caused by your code trying to convert from one type of pointer to another without indicating to the compiler that this is really what you want

This is what I'd do to remove the compilers doubt

Code:

#include <stdio.h>
int first (int(*q), int, int);
int second (int(*q)[4], int, int);
int main()
{
    int s[3][4]= {{1, 2, 3, 4},
                  {5, 6, 7, 8},
                  {9, 10, 11, 12}
                 };
    first (s[0], 3, 4);
    second (s, 3, 4);
    return 0;
}
int first (int *q, int x, int y)
{
    int i, j;
    for (i=0; i<x; i++)
    {
        for (j=0; j<y; j++)
            printf ("%d ", *(q+i*y+j));
        printf ("\n");
    }
    printf ("\n");
    return (*(q+i*y+j));
}
int second (int (*q)[4], int x, int y)
{
    int i, j, *p;
    for (i=0; i<x; i++)
    {
        p=q[i];
        for (j=0; j<y; j++)
            printf ("%d ", p[j]);
        printf ("\n");
    }
    return (*(p+j));
}

your code is trying to play the fact that pointer to the array and pointer to the first element of the array have same values (while not same type)

*(q+i*y+j) - is a result of "flattening" of the 2D array into 1D array. This is pointer arithmetic that lies behind the simple expression s[i][j]

PS. Noted that your first and second functions access out-of-bounds array member in their return statements

Originally Posted by Satya

The way you are passing the arguments to the first and second function are wrong. For the first function you have to pass it type int * and for the second of type int (*)[4]. Hence it is

Code:

first (&s[0][0], 3, 4);
second (&s[0], 3, 4);

You are mistaken for the call to the function named second: since an array is converted to a pointer to its first element in this context, this:

Code:

second(s, 3, 4);

is equivalent to:

Code:

second(&s[0], 3, 4);